IB Physics: Radioactive Decay Law and Half-Life Calculations | IB物理：放射性衰变规律与半衰期计算

📚 IB Physics: Radioactive Decay Law and Half-Life Calculations | IB物理：放射性衰变规律与半衰期计算

The study of radioactive decay is central to IB Physics. It explains how unstable nuclei transform, how we quantify the rate of decay, and how half-life serves as a practical measure for dating and medical applications. This article covers the decay law, decay constant, half-life calculations, and common IB exam pitfalls.

放射性衰变是IB物理课程中的核心内容之一，涉及不稳定原子核的转变、衰变速率的量化，以及半衰期在考古测年和医学中的应用。本文将系统讲解衰变规律、衰变常数、半衰期计算，并帮助同学们避开IB考试中的常见陷阱。

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1. The Nature of Radioactive Decay | 放射性衰变的本质

Radioactive decay is a random and spontaneous process. Each unstable nucleus has a certain probability of decaying per unit time, but we cannot predict exactly when a particular nucleus will decay. This randomness is key to understanding the statistical nature of decay.

放射性衰变是一个随机且自发的过程。每个不稳定原子核在单位时间内都有一定的衰变概率，但我们无法精确预测某个核在何时衰变。这种随机性是理解衰变统计规律的关键。

• Random: The time of decay of any single nucleus cannot be predicted.

  随机性：无法预测单个原子核的衰变时刻。

• Spontaneous: Decay is not affected by external factors such as temperature, pressure, or chemical state.

  自发性：衰变不受温度、压力或化学状态等外部因素影响。

• Statistical: With a large number of nuclei, the average behaviour is predictable.

  统计性：对于大量原子核，其平均行为是可预测的。

The number of decays per unit time is proportional to the number of undecayed nuclei present. This leads directly to the radioactive decay law.

单位时间内的衰变数与当前未衰变的原子核数目成正比，由此可直接导出放射性衰变规律。

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2. The Decay Law: N = N₀e^(−λt) | 衰变规律：N = N₀e^(−λt)

The decay law states that the rate of decay is proportional to the number of radioactive nuclei present. Mathematically, we write the differential equation:

衰变规律表明衰变速率与当前放射性原子核的数目成正比。用微分方程表示为：

dN/dt = −λN

where N is the number of undecayed nuclei, t is time, and λ (lambda) is the decay constant, which has units of s⁻¹ or a⁻¹. The negative sign indicates that N decreases over time.

其中N为未衰变的原子核数目，t为时间，λ（lambda）为衰变常数，单位是s⁻¹或a⁻¹。负号表示N随时间减少。

Solving this differential equation gives the exponential decay law:

解该微分方程得到指数衰减规律：

N = N₀e^(−λt)

where N₀ is the initial number of radioactive nuclei at t = 0. The same equation applies to activity A, with A = A₀e^(−λt), and to mass m, with m = m₀e^(−λt).

其中N₀是t = 0时初始放射性原子核数。该方程同样适用于活度A：A = A₀e^(−λt)，以及质量m：m = m₀e^(−λt)。

Be careful: the decay constant λ is different from half-life T₁/₂, though they are related. Do not confuse λ with the wavelength symbol.

注意：衰变常数λ与半衰期T₁/₂不同，但两者密切相关。切勿将λ与波长符号混淆。

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3. Half-Life: T₁/₂ = ln2 / λ | 半衰期：T₁/₂ = ln2 / λ

Half-life (T₁/₂) is the time taken for the number of radioactive nuclei to reduce to half of its original value. It is a constant for a given isotope under all normal conditions.

半衰期（T₁/₂）是放射性原子核数目减少到原来一半所需的时间。在通常条件下，特定同位素的半衰期是一个常量。

From the decay law, when N = N₀/2, we have:

由衰变规律，当N = N₀/2时：

N₀/2 = N₀e^(−λT₁/₂)

Taking natural logs on both sides:

两边取自然对数：

T₁/₂ = ln2 / λ

Since ln2 ≈ 0.693, an alternative form is T₁/₂ = 0.693/λ. This relationship is essential for converting between half-life and decay constant.

因为ln2 ≈ 0.693，所以也可写成T₁/₂ = 0.693/λ。该关系在计算中经常用于半衰期与衰变常数之间的转换。

Worked example: The half-life of iodine-131 is 8.02 days. What is its decay constant in s⁻¹?

例题：碘-131的半衰期为8.02天，求其衰变常数（单位s⁻¹）。

λ = 0.693 / (8.02 × 24 × 3600) = 9.99 × 10⁻⁷ s⁻¹

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4. Activity and Its Units | 活度及其单位

Activity A is the rate at which decays occur, defined as A = |dN/dt| = λN. The SI unit of activity is the becquerel (Bq), where 1 Bq = 1 decay per second.

活度A是单位时间内发生衰变的次数，定义为A = |dN/dt| = λN。活度的SI单位是贝克勒尔（Bq），1 Bq = 1次衰变每秒。

Since N decreases exponentially, activity also decreases exponentially:

由于N呈指数减少，活度也呈指数减小：

A = A₀e^(−λt)

where A₀ = λN₀.

其中A₀ = λN₀。

Worked example: A sample of strontium-90 has an initial activity of 500 Bq and a decay constant of 7.85 × 10⁻¹⁰ s⁻¹. How many strontium-90 nuclei are initially present?

例题：某锶-90样品初始活度为500 Bq，衰变常数为7.85 × 10⁻¹⁰ s⁻¹，求初始核数。

N₀ = A₀/λ = 500 / (7.85 × 10⁻¹⁰) = 6.37 × 10¹¹

Always check units: if A is in Bq and λ in s⁻¹, then N is a pure count.

务必检查单位：若A以Bq为单位，λ以s⁻¹为单位，则N为纯计数个数，无量纲。

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5. Graphical Representation | 图形表示

Exponential decay is represented by a curve that falls steeply at first, then approaches zero asymptotically. On a graph of N versus t, the half-life is the horizontal distance between points where N is halved.

指数衰减曲线先急剧下降，然后逐渐趋近于零。在N-t图中，半衰期是使N减半的相邻两点之间的时间间隔。

There is a very useful property: if you plot the natural logarithm of N (or A) against t, you get a straight line with slope −λ.

有一个非常有用的性质：若绘制ln N（或ln A）与t的关系图，可得一条斜率为−λ的直线。

ln N = ln N₀ − λt

This linear form is often used in exam data analysis questions. The y-intercept is ln N₀ and the slope is −λ.

该线性形式常用于数据分析题。y轴截距为ln N₀，斜率为−λ。

In an IB exam, you may be asked to determine the half-life from a graph. Read the time corresponding to half the initial count rate from the curve. For a logarithmic graph, use the slope to find λ, then T₁/₂ = 0.693/λ.

在IB考试中，可能要求从图中确定半衰期。从曲线中读取计数率降为初始值一半所对应的时间。对于对数坐标图，则利用斜率求λ，再用T₁/₂ = 0.693/λ。

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6. Calculating with Whole Numbers of Half-Lives | 用整数倍半衰期计算

When the elapsed time is an exact multiple of the half-life, calculations are simplest. After n half-lives, the remaining fraction is (1/2)ⁿ.

当经过时间为半衰期的整数倍时，计算最为简单。经过n个半衰期后，剩余比例为(1/2)ⁿ。

Remaining fraction = (1/2)ⁿ, where n = t / T₁/₂.

剩余比例 = (1/2)ⁿ，其中n = t / T₁/₂。

For example, after 3 half-lives, the remaining fraction is (1/2)³ = 1/8. After 5 half-lives, it is 1/32.

例如，经过3个半衰期后，剩余比例为(1/2)³ = 1/8。经过5个半衰期后，则为1/32。

Worked example: A sample contains 2.4 × 10¹⁰ radioactive nuclei with a half-life of 6 hours. How many remain after 24 hours?

例题：某样品含2.4 × 10¹⁰个放射性原子核，半衰期为6小时。24小时后还剩多少？

n = 24/6 = 4, so N = 2.4 × 10¹⁰ × (1/2)⁴ = 1.5 × 10⁹

This method avoids the exponential calculation entirely, but it only works for whole-number multiples. For arbitrary times, use the decay law directly.

这种方法完全避免了指数计算，但只能用于整数倍的情况。对于任意时间，应直接使用衰变规律。

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7. Worked IB-Style Problems | IB典型例题精讲

Problem 1: The activity of a radioactive sample falls from 800 Bq to 100 Bq in 9 days. Find the half-life.

例题1：某放射性样品活度从800 Bq降为100 Bq用了9天，求半衰期。

Solution: 800 → 400 → 200 → 100 is three half-lives in 9 days. Therefore T₁/₂ = 9/3 = 3 days.

解：800 → 400 → 200 → 100 经历了3个半衰期，共9天。因此T₁/₂ = 9/3 = 3天。

Problem 2: The half-life of carbon-14 is 5730 years. A sample of living wood has an activity of 0.25 Bq per gram. An ancient wooden artifact shows an activity of 0.0625 Bq per gram. Estimate the age of the artifact.

例题2：碳-14的半衰期为5730年。活木材每克活度为0.25 Bq，某古代木制文物每克活度为0.0625 Bq，估算文物的年代。

Solution: 0.25 → 0.125 → 0.0625 is two half-lives. Age = 2 × 5730 = 11460 years.

解：0.25 → 0.125 → 0.0625 经历了两个半衰期。年代 = 2 × 5730 = 11460年。

Problem 3: A radioactive source contains 4.0 × 10¹⁵ nuclei with a decay constant of 1.2 × 10⁻⁸ s⁻¹. Calculate the initial activity and the activity after 24 hours.

例题3：某放射源含4.0 × 10¹⁵个原子核，衰变常数为1.2 × 10⁻⁸ s⁻¹，求初始活度及24小时后的活度。

Initial activity: A₀ = λN₀ = 1.2 × 10⁻⁸ × 4.0 × 10¹⁵ = 4.8 × 10⁷ Bq.

初始活度：A₀ = λN₀ = 1.2 × 10⁻⁸ × 4.0 × 10¹⁵ = 4.8 × 10⁷ Bq。

After 24 h: A = 4.8 × 10⁷ × e^(−1.2 × 10⁻⁸ × 86400) = 4.8 × 10⁷ × e^(−0.0010368) ≈ 4.795 × 10⁷ Bq.

24小时后：A = 4.8 × 10⁷ × e^(−1.2 × 10⁻⁸ × 86400) = 4.8 × 10⁷ × e^(−0.0010368) ≈ 4.795 × 10⁷ Bq。

Note that the activity

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